Find $I=\oint_{C}[(e^{-x^{2}/2}-yz)\,dx+(e^{-y^{2}/2}+xz+2x) \,dy+(e^{-z^{2}/2}+5)\,dz]$, where $C$ is the circle $x=\cos t$, $y=\sin t$, $z=2$, $0\leq t\leq 2\pi$.

Solution  Let $\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}+R\,\mathbf{k}$, with $P=e^{-x^{2}/2}-yz$, $\ Q=e^{-y^{2}/2}+xz+2x$, and $R=e^{-z^{2}/2}+5$. It can be verified that $${\rm{curl }}\,{\bf{F}} = - x{\bf{i}} - y{\bf{j}}{\rm{ + (2 + 2}}z){\bf{k}}$$

To use Stokes' Theorem, take $S$ to be a plane region enclosed by the circle $C$ in the plane $z\rm{=2}$ so that $\mathbf{n}=\mathbf{k}$. Then $\rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}=6$ on $S$, and $$I=\iint\limits_{\kern-11pt S}{\rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}}\,dS=6\iint \limits_{\kern-11pt S}\,dx\,dy=6\,\pi$$